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What does Module(__name__) do?


What does Module(__name__) do?

By : user2953920
Date : November 21 2020, 07:31 AM
fixed the issue. Will look into that further
Within a module, the module’s name (as a string) is available as the value of the global variable __name__.
code :
>>> import itertools as it
>>> it.__name__
'itertools'
>>> dir(itertools)
['__doc__', '__name__', '__package__', 'chain', 'combinations', 'combinations_with_replacement', 'compress', 'count', 'cycle', 'dropwhile', 'groupby', 'ifilter', 'ifilterfalse', 'imap', 'islice', 'izip', 'izip_longest', 'permutations', 'product', 'repeat', 'starmap', 'takewhile', 'tee']


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What are the java equivalents of python's __file__, __name__, and Object.__class__.__name__?

What are the java equivalents of python's __file__, __name__, and Object.__class__.__name__?


By : LumpyTheCactus
Date : March 29 2020, 07:55 AM
help you fix your problem this.getClass() = current class
this.getClass().getPackage() = current package
Calling if __name__ == '__main__': in one module from a function in another module

Calling if __name__ == '__main__': in one module from a function in another module


By : user2600734
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , The if __name__ == '__main__' is mainly used to make a single python script executable. For instance, you define a function that does something, you use it by importing it and running it, but you also want that function to be executed when you run your python script with python module1.py.
For the question you asked, the best I could come up with is that you wanted a function defined in "module1.py" to run when you invoke "module2.py". That would be something like this:
code :
### module1.py:
def main():
    # does something
    ...

if __name__ == '__main__':
    main()

### module2.py:
from module1 import main as main_from_module_one

if __name__ == '__main__':
    main_from_module_one()  # calling function main defined in module1
The 'class object' prints its name using `.__name__`,despite of the absence of `__name__` attribute

The 'class object' prints its name using `.__name__`,despite of the absence of `__name__` attribute


By : Elad Undler
Date : March 29 2020, 07:55 AM
This might help you After the class body is executed Python will fill in some attributes automatically. That includes __name__ but also __doc__, __qualname__ (Python 3.4+) and __module__. The complete list of these automated attributes is avaiable as table in the inspect module documentation:
code :
Type    Attribute       Description
class   __doc__         documentation string
        __name__        name with which this class was defined
        __qualname__    qualified name
        __module__      name of module in which this class was defined
>>> '__name__' in dir(object.__class__)
True
Dynamically changing a function __name__ throws AttributeError: 'method' object has no attribute '__name__'

Dynamically changing a function __name__ throws AttributeError: 'method' object has no attribute '__name__'


By : ssrulz1
Date : March 29 2020, 07:55 AM
wish of those help The reason why handler.__name__ = name doesn't work is because handler is a bound method, i.e. it's an object that encapsulates a reference to a function and the instance that'll take the value of self. You can confirm this by printing it out; it says "bound method":
code :
>>> a.imaginary_method
<bound method A.__global_handler of <__main__.A object at 0x00000150E64FC2B0>>
def __get_global_handler(self, name):
    handler = self.__global_handler
    handler.__func__.__name__ = name # Change the method's name
    handler.__func__.__qualname__ = __class__.__qualname__ + '.' + name
    return handler
a = A()
foo = a.foo
bar = a.bar

print(foo)  # <bound method A.bar of <__main__.A object at 0x00000212AFCC7198>>
print(bar)  # <bound method A.bar of <__main__.A object at 0x00000212AFCC7198>>
class MethodFactory:
    def __getattr__(self, name):
        def func(*args, **kwargs):
            print("I am an imaginary method with name", name)
            print("My arguments are:", args)
            print("My keyword arguments are:", kwargs)

        func.__name__ = name
        func.__qualname__ = __class__.__qualname__ + '.' + name
        return func

a = MethodFactory()
foo = a.foo
bar = a.bar
print(foo)  # <function MethodFactory.foo at 0x00000238BDADC268>
print(bar)  # <function MethodFactory.bar at 0x00000238BF8AB730>
class MethodFactory:
    def __getattr__(self, name):
        def func(self, *args, **kwargs):
            print("I am an imaginary method with name", name)
            print("My arguments are:", args)
            print("My keyword arguments are:", kwargs)

        func.__name__ = name
        func.__qualname__ = __class__.__qualname__ + '.' + name
        return func.__get__(self, type(self))

a = MethodFactory()
foo = a.foo
bar = a.bar
print(foo)  # <bound method MethodFactory.foo of <__main__.MethodFactory object at 0x00000137BF0C29E8>>
print(bar)  # <bound method MethodFactory.bar of <__main__.MethodFactory object at 0x00000137BF0C29E8>>
Import in a module fails because __name__ is __main__

Import in a module fails because __name__ is __main__


By : user3603835
Date : March 29 2020, 07:55 AM
may help you . All you should have to do is change your import in main.py to from myPackage.subfolder2 import script2. I set up a directory and some files in this way, using that import, and the script runs as expected:
code :
main.py
myPackage/
  subfolder1/
    script11.py
  subfolder2/
    script2.py
def bar():
    return 10
from ..subfolder1.script11 import bar

def foo():
    x = bar()
    print('foo is', x)
from myPackage.subfolder2 import script2 as s2

s2.foo()
>>> py .\main.py
foo is 10
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